Answer
$$h'(u) = \frac{\sqrt{u}}{u+1}$$
Work Step by Step
$$h(u) = \int\limits_0^u {\frac{\sqrt{t}}{t+1}}dt$$
Plug in the upper bound into t and multiply by the derivative of the upper bound due to chain rule.
$$h'(u) = \frac{\sqrt{u}}{u+1} \times (u)'$$
Simplify.
$$h'(u) = \frac{\sqrt{u}}{u+1}$$
